As for exponential stability, there exists a characterization for a \(\mathcal {C}_{0}\) -semigroup of contraction to be polynomially stable: Theorem 2.4 in [1]}. Taking into account that \(\mathbf {i}\mathbb {R}\subset \rho (\mathcal {A}_2)\) , The polynomial stability follows from the estimate
\(\limsup _{\beta \in \mathbb {R}, \left| \beta \right| \rightarrow +\infty } \frac{1}{\beta ^2}\left\Vert  (\mathbf {i}\beta I-A)^{-1}\right\Vert  _{\mathcal {L}(X)}<\infty , \) 
Suppose the estimate (REF ) is false. Then there exists a sequence \((w_{n})\)  of real numbers, with \(|w_{n}|\longrightarrow +\infty \)  (without loss of generality, we suppose that \(w_n>0\) ), and a sequence of vectors \((y_{n})=(u_{n},v_{n},\theta _{2,n},q_{2,n})\)  in \(\mathcal {D}(\mathcal {A}_2)\)  with \(\left\Vert  y_{n}\right\Vert  _{\mathcal {H}_2}=1\) , such that
\(\left\Vert  w_{n}^{2}(\mathbf {i}w_{n}I-\mathcal {A}_2)y_{n}\right\Vert  _{\mathcal {H}_2}\longrightarrow 0\) 
which is equivalent to
\(w_{n}^{2}\left( \mathbf {i}w_{n}u_{1,n}-v_{1,n}\right)&=&f_{1,n}\longrightarrow 0,\;\;\;\text{in}\;H^{1}(0,\ell _{1}), \\w_{n}^{2}\left( \mathbf {i}w_{n}v_{1,n}-\alpha _{1}\partial _{x}^{2}u_{1,n}\right) &=&g_{1,n}\longrightarrow 0,\;\;\;\text{in}\;L^{2}(0,\ell _{1}), \) 
and
\(w_{n}^{2}\left( \mathbf {i}w_{2,n}u_{2,n}-v_{2,n}\right)&=&f_{2,n}\longrightarrow 0,\;\;\;\text{in}\;H^{2}(0,\ell _{2}), \\w_{n}^{2}\left( \mathbf {i}w_{n}v_{2,n}+\partial _x^2\left(\alpha _{2}\partial _{x}^{2}u_{2,n}-\beta _{2}\theta _{2,n}\right)\right)&=&g_{2,n}\longrightarrow 0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}), \\w_{n}^{2}\left( \mathbf {i}w_{n}\theta _{2,n}+\gamma _{2}\partial _{x}^{2}v_{2,n}+\delta _{2}\partial _{x}q_{2,n}\right)&=&h_{2,n}\longrightarrow 0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}), \\w_{n}^{2}\left( \mathbf {i}w_{n}\tau _{2}q_{2,n}+q_{2,n}+\kappa _{2}\partial _{x}\theta _{2,n}\right) &=&d_{2,n}\longrightarrow 0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}), \) 
We will prove that \(y_n\)  converges to zero in \(\mathcal {H}_2,\)  which contradict the fact that \(\left\Vert  y_{n}\right\Vert  _{\mathcal {H}_2}=1.\)
First, since
\(Re(\left\langle w_{n}^{2}(\mathbf {i}w_{n}-\mathcal {A}_2)y_{n},y_{n}\right\rangle _{\mathcal {H}_2})=-\frac{\gamma _{2}}{\kappa _{2}}w_{n}^{2}\left\Vert  q_{2,n}\right\Vert  ^{2},\) 
we obtain that \(w_{n}q_{2,n}\)  converges to 0 in \(L^{2}(0,\ell _{2})\) , and then by (), we deduce that \(\partial _{x}\theta _{2,n}\)  converges to zero in \(L^{2}(0,\ell _{2}).\)
Now, from (REF )-() we deduce
\(w_{n}^{2}\left( w_{n}^{2}u_{1,n}+\alpha _{1}\partial _{x}^{2}u_{1,n}\right) &=&-g_{1,n}-\mathbf {i}w_{n}f_{1,n},  \\w_{n}^{2}\left( -w_{n}^{2}u_{2,n}+\partial _x^2\left(\alpha _{2}\partial _{x}^{2}u_{2,n}-\beta _{2}\theta _{2,n}\right)\right) &=&g_{2,n}+\mathbf {i}w_{n}f_{2,n},  \\w_{n}^{2}\left( \theta _{2,n}+\frac{1}{\mathbf {i}w_{n}}\gamma _{2}\partial _{x}q_{2,n}+\delta _{2}\partial _{x}^{2}u_{2,n}\right) &=&\frac{1}{\mathbf {i}w_{n}}(h_{2,n}+\partial _{x}^{2}f_{2,n}) \) 
and
\(w_{n}^{2}\left\Vert  v_{j,n}\right\Vert  ^{2}-w_{n}^{4}\left\Vert  u_{j,n}\right\Vert ^{2}\longrightarrow 0,\;\;j=1,2.\) 
We divide the rest of the proof into three steps:
First step: We prove some lemmas that will be used frequently later:
Let \(p_2(x)=ax+b\) , with (\(a,b\in \mathbb {C}\) ), taking the real part of the inner product of (), () and () by \(p_2w_n^{\beta -2}\partial _{x}u_{2,n},\)  \(p_2w_n^{\beta -2}\partial _{x}\theta _{2,n}\)  and \(p_2w_n^{\beta -2}\frac{1}{\mathbf {i}w_n}\partial _{x}q_{2,n}\)  respectively (in \(L^2(0,\ell _2)\) ), where \(\beta \)  is a non negative real number such that \(\beta \le 1\) , we obtain
\(-\left. \frac{1}{2} w_{n}^{2+\beta }\left| u_{2,n}(x)\right| ^{2}p_2(x)\right]_0^{\ell _2} +\frac{1}{2}\int _{0}^{\ell _{2}}w_{n}^{2+\beta }\left| u_{2,n}\right|^{2}\partial _xp_2 dx-\frac{1}{2}\left. w_n^\beta \alpha _2 \left| \partial _{x}^2u_{2,n}(0)\right| ^{2}p_2(x)\right] _0^{\ell _2}\\+\left. w_{n}^{\beta }\partial _x u_{2,n}(x)\left(\alpha _{2}\partial _{x}^{3}u_{2,n}(x)-\beta _{2}\partial _x\theta _{2,n}(x) \right) p_2(x)\right]_0^{\ell _2} +\frac{3}{2}w_n^\beta \alpha _2\int _{0}^{\ell _{2}}\left| \partial _{x}^2u_{2,n}\right|^{2}\partial _xp_2dx \\-Re\left( w_n^\beta \beta _{2}\left\langle \theta _{2,n},\partial _{x}^2u_{2,n}\partial _x p_2 \right\rangle +w_n^\beta \beta _{2}\left\langle \partial _{x}\theta _{2,n},\partial _{x}^{2}u_{2,n}p_2\right\rangle \right)=\circ (1),\\\frac{1}{2}w_n^\beta \int _{0}^{\ell _{2}}\left| \theta _{2,n}\right|^{2}\partial _xp_2dx- Re\left(\textbf {i} w_n^{\beta -1}\gamma _{2}\left\langle p_2\partial _{x}\theta _{2,n},\partial _{x}q_{2,n}\right\rangle \right) \\-\delta _2 w_n^\beta Re\left\langle \partial _{x}\theta _{2,n},p_2\partial _{x}^2u_{2,n}\right\rangle =\circ (1) \) 
and
\(&&-\frac{1}{2}\left. w_n^\beta \tau _2 \left| q_{2,n}(x)\right| ^{2}p_2(x)\right] _0^{\ell _2}+ Re\left(\textbf {i} w_n^{\beta -1}\kappa _2\left\langle p_2\partial _{x}\theta _{2,n},\partial _{x}q_{2,n}\right\rangle \right)=\circ (1). \) 
We have used that \(\theta _{2,n}(0)=\theta _{2,n}(\ell )=0\)  and that \(w_nq_n=o(1)\) .
Multiplying (REF ) by \(\delta _2\) , () by \(\beta _2\) , and (REF ) by \(\frac{\beta _2 \gamma _2}{\kappa _2}\) , then summing, we get, by taking \(p_2(x)=\ell _2-x\) ,
Lemma 3.3
\(\frac{\ell _2}{2}\delta _2 w_{n}^{2+\beta }\left| u_{2,n}(0)\right| ^{2}+\frac{\ell _2}{2}w_n^\beta \alpha _2 \delta _2 \left| \partial _{x}^2u_{2,n}(0)\right| ^{2}+\frac{\ell _2}{2}\frac{\beta _2 \gamma _2}{\kappa _2}w_n^\beta \tau _2 \left| q_{2,n}(0)\right| ^{2}\\-\frac{1}{2}\delta _2\int _{0}^{\ell _{2}}w_{n}^{2+\beta }\left| u_{2,n}\right|^{2}dx -\frac{3}{2}w_n^\beta \alpha _2 \delta _2\int _{0}^{\ell _{2}}\left| \partial _{x}^2u_{2,n}\right|^{2}dx-\frac{1}{2}\beta _2 w_n^\beta \int _{0}^{\ell _{2}}\left| \theta _{2,n}\right|^{2}dx \\+Re \left( w_n^\beta \beta _{2} \delta _2\left\langle \theta _{2,n},\partial _{x}^2u_{2,n}\right\rangle \right).=\circ (1), \) 

The second lemma gives \(\alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\)  in terms of \(\partial _{x}^{2}u_{2,n}(0)\) .
Lemma 3.4 
For every non negative \(\beta \le \frac{1}{2}\)  there exists a positive number \(a\)  such that
\(\frac{w_n^\beta }{w_{n}^{1/2}}(\alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0))+w_n^\beta \alpha _{2}a\partial _{x}^{2}u_{2,n}(0)+w_n^{1+\beta }(\alpha _{2}+\beta _2 \delta _2)a^{3} u_{2,n}(0)=o(1).\) 

Let \(a\in \mathbb {R}\) . The inner product of \(\alpha _{2}\partial _{x}^{4}u_{2,n}-\beta _2\partial _{x}^{2}\theta _{2,n}\)  with \(\frac{w_n^\beta }{w_{n}^{1/2}}e^{-aw_{n}^{1/2}x}\)  gives
\(w_n^\beta \left\langle \alpha _{2}\partial _{x}^{2}\left( \partial _{x}^{2}u_{2,n}-\beta _2\theta _{2,n}\right) ,\frac{1}{w_{n}^{1/2}}e^{-aw_{n}^{1/2}x}\right\rangle =w_n^\beta \left( \frac{1}{w_{n}^{1/2}}\left[\left( \alpha _{2}\partial _{x}^{3}u_{2,n}-\beta _2\partial _{x}^{2}\theta _{2,n}\right) e^{-aw_{n}^{1/2}x}\right] _{0}^{\ell _{2}}\right. \\+\alpha _{2}a\left[ e^{-aw_{n}^{1/2}x}\partial _{x}^{2}u_{2,n}\right] _{0}^{\ell _{2}}+\alpha _{2}a^{2}w_{n}^{1/2}\left[ e^{-aw_{n}^{1/2}x}\partial _{x}u_{2,n}\right] _{0}^{\ell _{2}}+\alpha _{2}a^{3}w_{n}\left[e^{-aw_{n}^{1/2}x}u_{2,n}\right] _{0}^{\ell _{2}}\\\left. +\alpha _{2}a^{4}w_{n}^{3/2}\left\langle u_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle -\beta _{2}a\left[ e^{-aw_{n}^{1/2}x}\theta _{2,n}\right] _{0}^{\ell _{2}}-\beta _{2}a^{2}w_{n}^{1/2}\left\langle \theta _{2,n},e^{-aw_{n}^{1/2}x}\right\rangle \right).\) 
The inner product of () by \(w_{n}^{\frac{1}{2}+\beta -2}e^{-aw_{n}^{1/2}x}\)  gives
\(&&-w_{n}^{\beta +1/2}\left\langle \theta _{2,n},e^{-aw_{n}^{1/2}x}\right\rangle \\&=& -\mathbf {i} w_{n}^{\beta -1/2}\gamma _{2}\left\langle \partial _{x}q_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle +\delta _{2}w_{n}^{\beta +1/2}\left\langle \partial _{x}^{2}u_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle -\frac{w_{n}^{\frac{1}{2}+\beta -2}}{\mathbf {i}w_{n}^{1/2}}\left\langle h_{2,n}+\partial _{x}^{2}f_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle \\&=& -\mathbf {i}w_{n}^{\beta -1/2}\gamma _{2}\left[ e^{-aw_{n}^{1/2}x}q_{2,n}\right]_{0}^{\ell _{2}}-\mathbf {i}\gamma _{2}w_n^\beta a\left\langle q_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle +\delta _{2}w_{n}^{1/2+\beta }\left[e^{-aw_{n}^{1/2}x}\partial _xu_{2,n}\right] _{0}^{\ell _{2}} \\&+&\delta _{2}aw_{n}^{1+\beta }\left[e^{-aw_{n}^{1/2}x}u_{2,n}\right] _{0}^{\ell _{2}}+\delta _{2}a^{2}w_{n}^{\beta +3/2}\left\langle u_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle +o(1)\) 
It yields that the inner product of () with \(w_{n}^{-\frac{1}{2}+\beta -2}e^{-aw_{n}^{1/2}x}\)  gives
\(&&+((\alpha _{2}+\beta _{2}\delta _{2})a^{4}-1)w_{n}^{3/2+\beta }\left\langle u_{2,n},e^{-aw_{n}^{1/2}x}\right\rangle +\frac{w_n^\beta }{w_{n}^{1/2}}\left[\left( \alpha _{2}\partial _{x}^{3}u_{2,n}-\beta _2\partial _{x}^{2}\theta _{2,n}\right) e^{-aw_{n}^{1/2}x}\right] _{0}^{\ell _{2}}\\&&+w_n^\beta \alpha _{2}a\left[ e^{-aw_{n}^{1/2}x}\partial _{x}^{2}u_{2,n}\right] _{0}^{\ell _{2}}+w_n^{\beta +1/2}(\alpha _{2}+\beta _2 \delta _2)a^{2}\left[ e^{-aw_{n}^{1/2}x}\partial _{x}u_{2,n}\right] _{0}^{\ell _{2}}\\&& +w_n^{1+\beta }(\alpha _{2}+\beta _2 \delta _2)a^{3}\left[e^{-aw_{n}^{1/2}x}u_{2,n}\right] _{0}^{\ell _{2}}-w_n^\beta \beta _{2}a\left[ e^{-aw_{n}^{1/2}x}\theta _{2,n}\right] _{0}^{\ell _{2}} =o(1)\) 
Then, (REF ) is immediate by choosing \(a=\frac{1}{(\alpha _{2}+\beta _{2}\delta _{2})^{1/4}}\) .      \(\sqcup \unknown. {\sqcap }\)
Using the boundedness of \(y_n\)  in Lemma REF  with \(\beta =0\) , we deduce that
Lemma 3.5
\(w_nu_{2,n}(0)=O(1)\;\;\;\text{and}\;\;\;\partial _{x}^{2}u_{2,n}(0)=O(1), \) 
then, from Lemma REF  we have
\(\frac{1}{w_n^{1/2}}\left( \alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\right) =O(1).\) 

Lemma 3.6 For every \(0\le \beta \le 1\)  we have
\(w_{n}^{\beta }\left[ \left\Vert  \theta _{2,n}\right\Vert ^{2}+\delta _{2}\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2} -2\delta _{2}Re\left\langle \partial _{x} u_{2,n},\partial _{x}\theta _{2,n}\right\rangle \right] =o(1) \) 

Multiplying () by \(w_{n}^{\beta -2}\theta _{2,n}\)  and \(\delta _2w_{n}^{\beta -2}\partial _{x}^{2}u_{2,n}\)  respectively, we get
\(w_{n}^{\beta }\left\Vert  \theta _{2,n}\right\Vert ^{2}+\mathbf {i} w_{n}^{\beta -1}\gamma _2\left\langle q_n,\partial _x\theta _{2,n}\right\rangle +\delta _{2}w_{n}^{\beta }\left\langle \partial ^2_{x} u_{2,n},\theta _{2,n}\right\rangle =o(1)\) 
and
\(&&\delta _{2}^2w_{n}^{\beta }\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2}+\delta _{2}w_{n}^{\beta }\left\langle \theta _{2,n},\partial _{x}^{2}u_{2,n}\right\rangle -\mathbf {i}w_{n}^{\beta -1}\delta _{2}\gamma _{2}\left\langle \partial _xq_{2,n},\partial _{x}^{2}u_{2,n}\right\rangle =o(1). \) 
Summing (REF ) with (REF ), we find
\(w_{n}^{\beta }\left\Vert  \theta _{2,n}\right\Vert ^{2}+\delta _{2}^2w_{n}^{\beta }\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2} +2\delta _{2}w_{n}^{\beta }Re \left\langle \partial ^2_{x} u_{2,n},\theta _{2,n}\right\rangle \\+\mathbf {i}w_{n}^{\beta -1} \gamma _2\left( \left\langle q_n,\partial _x\theta _{2,n}\right\rangle -\mathbf {i}\delta _{2}w_{n}^{\beta -1}\left\langle \partial _xq_{2,n},\partial _{x}^{2}u_{2,n}\right\rangle \right) =o(1) \) 
It yelds, after integrating by parts
\(w_{n}^{\beta }\left\Vert  \theta _{2,n}\right\Vert ^{2}+\delta _{2}^2w_{n}^{\beta }\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2} +2\delta _{2}w_{n}^{\beta }Re \left\langle \partial ^2_{x} u_{2,n},\theta _{2,n}\right\rangle \\+\mathbf {i}w_{n}^{\beta -1}\gamma _2\left[ \left\langle q_n,\partial _x\theta _{2,n}\right\rangle +\mathbf {i}\delta _{2}q_{2,n}(0)\partial _{x}^{2}\overline{u_{2,n}}(0)+\mathbf {i}\delta _{2}\left\langle q_{2,n},\partial _{x}^{3}u_{2,n}\right\rangle \right] =o(1).\) 
Using that \(w_{n}q_{2,n}, \partial _{x}\theta _{2,n} \longrightarrow 0\)  in \(L^{2}(0,\ell _{2})\) , we have
\(w_{n}\left\langle q_{2,n},\partial _{x}\theta _{2,n}\right\rangle \rightarrow 0.\) 
From () we deduce that \(\frac{\partial _x^2\left(\alpha _{2}\partial _{x}^{2}u_{2,n}-\beta _{2}\theta _{2,n}\right)}{w_{n}}\)  is bounded. Then, using the Gagliardo-Nirenberg inequality and the boundedness of \(\alpha _{2}\partial _{x}^{2}u_{2,n}-\beta _{2}\theta _{2,n}\) , we deduce that
\(\frac{\left\Vert \partial _x\left(\alpha _{2}\partial _{x}^{2}u_{2,n}-\beta _{2}\theta _{2,n}\right)\right\Vert  }{w_{n}^{1/2}}\)  is bounded.
Then, taking into account that \(\partial _{x}\theta _{2,n} \longrightarrow 0\)  in \(L^{2}(0,\ell _{2})\) , it yields, \(\frac{\left\Vert \alpha _2\partial _{x}^{3}u_{2,n}\right\Vert  }{w_{n}^{1/2}}=O(1)\) .
Hence,
\(w_{n}^{1/2}\left\langle q_{2,n},\partial _{x}^{3}u_{2,n}\right\rangle \rightarrow 0.\) 
(We have used that \(w_nq _{2,n} \longrightarrow 0\) ).
Now, using again that \(w_nq _{2,n} \longrightarrow 0\) , and that \(\frac{1}{w_n}\partial _xq_{2,n}=O(1)\)  (by the boundedness of \(\theta _{2,n}\)  and \(\partial _{x}^{2}u_{2,n}\) ), we can deduce, via the Gagliardo-Nirenberg inequality, that
\(q_{2,n}(0)=o(1). \) 
Using (REF ) and (REF ), we get
\(q_{2,n}(0)\partial _{x}^{2}\overline{u_{2,n}}(0)=o(1).\) 
Hence (REF ) holds true.      \(\sqcup \unknown. {\sqcap }\)
Second step: We prove that \(w_{n}^{1/2}\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2}\) , \(w_{n}^{1/2}\left\Vert  \theta _{2,n}\right\Vert ^{2}\)  and \(w_{n}^{1/2} \left\Vert  v_{2,n}\right\Vert  ^{2}\)  converge to zero.
By Gagliardo-Nirenberg inequality and the boundedness of \(w_nu_n\)  and \(\partial _x^2u_{2,n}\) , we have that \(w^{1/2}\partial _{x}u _{2,n}\)  is bounded. Recall again that \(\partial _{x}\theta _{2,n}=o(1)\) , then
\(w_n^{1/2}\left\langle \partial _{x} u_{2,n},\partial _{x}\theta _{2,n}\right\rangle =o(1)\) 
which implies, using (REF ) with \(\beta =1/2\) ,
\(w_{n}^{1/2}\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2}=o(1)\;\;\text{and}\;\;w_{n}^{1/2}\left\Vert  \theta _{2,n}\right\Vert ^{2}=o(1).\) 
Taking the inner product of () with \(w_{n}^{-1}u_{2,n}\) , we get
\(&&w_{n}^{3}\left\Vert  u_{2,n}\right\Vert  ^{2}=\alpha _{2}w_{n}\Vert \partial _{x}^{2}u_{2,n}\Vert ^2+\beta _{2}w_n\left\langle \partial _x\theta _{2,n},\partial _{x}u_{2,n}\right\rangle \\&-&w_{n}\overline{u_{2,n}}(0)\left( \alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\right) +o(1). \) 
Using (REF ) and (REF ), we have that
\(w_{n}^{1/2}u_{2,n}(0)\left( \alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\right) =O(1).\) 
and a fortiori
\(u_{2,n}(0)\left( \alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\right) =o(1).\) 
Using (REF ), (REF ) and (REF ) in (REF ) yields
\(w_{n}\left\Vert u_{2,n}\right\Vert =o(1).\) 
Now we take \(\beta =0\)  in (REF ) to deduce in particular that
\(w_nu_{2,n}(0)=o(1)\) 
which implies, keeping in mind (REF ),
\(w_{n}^{1/2}u_{2,n}(0)\left( \alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\right) =o(1).\) 
Return back to (REF ), it yields
\(w_{n}^{5/2}\left\Vert u_{2,n}\right\Vert ^{2}=o(1).\) 
Third step: We prove that \(w_{n}\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2}\) , \(w_{n}\left\Vert  \theta _{2,n}\right\Vert ^{2}\)  and \(w_{n} \left\Vert  v_{2,n}\right\Vert  ^{2}\)  converge to zero.
Using the estimates of the second step in (REF ) with \(\beta =1/2\) , we get
\(w_n^{1/4}q_n(0)=o(1),\;\;w_n^{1/4}\partial _x^2u_{2,n}(0)=o(1),\;\;w_n^{5/4}u_{2,n}(0)=o(1),\) 
and then by (REF ),
\(\frac{\alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)}{w_n^{1/4}}.\) 
We deduce in particular that
\(w_{n}u_{2,n}(0)\left( \alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)\right) =o(1).\) 
Now, by using again the Gagliardo-Nirenberg inequality, we have for \(\varepsilon >0\) ,
\(|w_{n}\left\langle \partial _{x} u_{2,n},\partial _{x}\theta _{2,n}\right\rangle |&\le & \frac{\varepsilon }{2}w_n^2\Vert \partial _{x} u_{2,n}\Vert ^2+\frac{1}{2\varepsilon }\Vert \partial _{x} \theta _{2,n}\Vert ^2\\&\le & C^2\varepsilon \left(w_n^{3/2}\Vert  u_{2,n}\Vert w_n^{1/2}\Vert \partial ^2_{x} u_{2,n}\Vert +w_n^{2}\Vert  u_{2,n}\Vert ^2 \right)+ \frac{1}{2\varepsilon }\Vert \partial _{x} \theta _{2,n}\Vert ^2\\&\le & \frac{C^2\varepsilon }{2}\left(w_n^{3}\Vert  u_{2,n}\Vert ^2+w_n\Vert \partial ^2_{x} u_{2,n}\Vert ^2+2w_n^{2}\Vert  u_{2,n}\Vert ^2 \right)+ \frac{1}{2\varepsilon }\Vert \partial _{x} \theta _{2,n}\Vert ^2\) 
Combining (REF ), with \(\beta =1\) , (REF ), (REF ) and (REF ), with \(\varepsilon \)  small enough we obtain that:
\(w_{n}\left\Vert \theta _{2,n}\right\Vert ^{2},\;\;w_{n}\left\Vert  \partial _{x}^{2}u_{2,n}\right\Vert ^{2},\;\;w_{n}^{3}\left\Vert  u_{2,n}\right\Vert ^{2}\longrightarrow 0.\) 
Fourth step: Conclusion.
Using (REF ) with \(\beta =1\)  we get
\(w_n^{1/2}q_n(0)=o(1),\;\;w_n^{1/2}\partial _x^2u_{2,n}(0)=o(1),\;\;w_n^{3/2}u_{2,n}(0)=o(1) \) 
and then by (REF ),
\(\alpha _{2}\partial _{x}^{3}u_{2,n}(0)-\beta _{2}\partial _{x}\theta _{2,n}(0)=o(1).\) 
Finally we prove that \(\left\Vert  y_{n}\right\Vert  _{\mathcal {H}}\longrightarrow 0:\)
To this end, taking the real part of the inner product of
() by \((\ell _1-x)w_n^{-2}\partial _{x}u_{1,n}\)  to get
\(\int _{0}^{\pi }\left( \left| \partial _{x}u_{1,n}(x)\right|^{2}+w_{n}^{2}\left| u_{1,n}(x)\right| ^{2}\right) dx\longrightarrow 0.\) 
We have used (REF ), (REF ), the continuity condition of \(u_{n}\)  at \(x=0\) , the boundary condition of \(u_n\)  at \(x=\ell _1\)  and the damping condition
at \(x=0\) , in (REF ). In summary, we have \(\left\Vert  y_{n}\right\Vert  _{\mathcal {H}}\longrightarrow 0.\)  Such result contradicts the hypothesis that \(y_{n}\)  has the unit norm.      \(\sqcup \unknown. {\sqcap }\)
Lack of exponential stability
For simplicity, we take, in this section, \(\ell _{1}=\ell _{2}=\pi \) .
Theorem 3.7
The system (\(\mathcal {S}_2\) ) is not
exponentially stable in the energy space \(\mathcal {H}_2\) . Moreover, the associated semigroup is polynomially stable at least for the order 1 i.e. for every \(\alpha < 1\)  we have :
\(\limsup _{|\lambda |\rightarrow \infty } |\lambda |^{-\alpha }\left\Vert (i\lambda I - \mathcal {A})^{-1}\right\Vert _{\mathcal {L}(\mathcal {H})}=\infty . \) 

To prove (REF ), it is enough to show that there exist a bounded
sequence \(f_{n}\)  in \(\mathcal {H}_2\)  and a sequence \(w_{n}\)  of real numbers
with \(w_{n}\rightarrow \infty \)  such that the solution \(y_{n}\)  of
\(w_n^\alpha \left( \mathbf {i}w_{n}-\mathcal {A}_2\right) y_{n}=f_{n} \) 
satisfies
\(\lim \limits _{n\rightarrow 0}\left\Vert  y_{n}\right\Vert  _{\mathcal {H}_2}=\infty . \) 
For each \(n\in \mathbb {N},\)  we choose \(f_{n}=(0,0,-\alpha _{1}\sin \frac{w_{n}}{\sqrt{\alpha _{1}}}x,0,0,0)\) . Then \(f_{n}\in \mathcal {H}_2\)  and
is bounded in \(\mathcal {H}_2.\)  The solution \(y_{n}\)  of (REF )
satisfies
\(w_n^\alpha \left( \mathbf {i}w_{n}u_{1,n}-v_{1,n}\right) &=&0,\;\;\;\text{in}\;H^{1}(0,\ell _{1}),  \\w_n^\alpha \left( \mathbf {i}w_{n}v_{1,n}-\alpha _{1}\partial _{x}^{2}u_{1,n}\right) &=&-\alpha _{1}\sin \frac{w_{n}}{\sqrt{\alpha _{1}}},\;\;\;\text{in}\;L^{2}(0,\ell _{1}) \) 
and
\(w_n^\alpha \left( \mathbf {i}w_{2,n}u_{2,n}-v_{2,n}\right) &=&0,\;\;\;\text{in}\;H^{2}(0,\ell _{2}),  \\w_n^\alpha \left( \mathbf {i}w_{n}v_{2,n}+\partial _{x}^{2}\left( \alpha _{2}\partial _{x}^{2}u_{2,n}-\beta _{2}\theta _{2,n}\right)\right) &=&0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}),  \\w_n^\alpha \left( \mathbf {i}w_{n}\theta _{2,n}+\delta _{2}\partial _{x}^{2}v_{2,n}+\gamma _{2}\partial _{x}q_{2,n}\right) &=&0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}), \\w_n^\alpha \left( \mathbf {i}w_{n}\tau _2q_{2,n}+q_{2,n}+\kappa _{2}\partial _{x}\theta _{2,n}\right) &=&0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}). \) 
Let \(\gamma _{n}:=\frac{w_{n}}{\sqrt{\alpha _{1}}}\)  and substitute equation (REF ) into () to obtain the second order equation
\(w_n^\alpha \left( \gamma _{n}^{2}u_{1,n}+\partial _{x}^{2}u_{1,n}\right)=\alpha _{1}\sin \gamma _{n}x ,\) 
which can be solved by
\(u_{1,n}=\frac{1}{w_n^\alpha }\left( c_{1}\sin (\gamma _{n}x)+(-\frac{x}{2\gamma _{n}}+c_{2})\cos (\gamma _{n}x)\right)\) 
where the complex numbers \(c_{1}\)  and \(c_{2}\)  depend on \(n\) .
Substitute (REF ) into () and () to get
\(-w_{n}^{2}u_{2,n}+\alpha _{2}\partial _{x}^{4}u_{2,n}-\beta _{2}\partial _{x}^{2}\theta _{2,n} &=&0,  \\\mathbf {i}w_{n}\theta _{2,n}+\mathbf {i}w_{n}\delta _{2}\partial _{x}^{2}u_{2,n}+\gamma _{2}\partial _{x}q _{2,n} &=&0. \) 
Notice that, by (), we have \(\partial _{x}\theta _{2,n}\in H^1(0,\pi )\) .
We replace \(\partial _xq_{2,n}\)  in () by \(-\frac{\kappa _2}{\mathbf {i}w_n\tau _2+1}\partial _x^2\theta _{2,n},\)  then multiplying the obtained result by \(\mathbf {i}w_n\)  to get
\(\mathbf {i}w_n\theta _{2,n}-\frac{\gamma _2 \kappa _2}{\mathbf {i}w_n\tau _2+1}\partial _{x}^{2}\theta _{2,n}+\mathbf {i}w_n\delta _{2}\partial _{x}^{2}u_{2,n} =0,\;\;\;\text{in}\;L^{2}(0,\ell _{2}). \) 
Combining (REF ) and (REF ), it yields
\(\alpha _{2} \kappa _{2} \gamma _2 \partial _{x}^{6}u_{2,n}-\mathbf {i}w_n(\mathbf {i}w_n\tau _2+1)(\alpha _{2}+\delta _2\beta _{2})\partial _{x}^{4}u_{2,n}-\gamma _2\kappa _{2}w_{n}^{2}\partial _{x}^{2}u_{2,n}+\mathbf {i}w_n^3(\mathbf {i}w_n\tau _2+1)u_{2,n}=0 \) 
which is an ordinary differential equation of order 6 in \(u_{2,n}.\)  Its
characteristic equation is
\(\alpha _{2} \kappa _{2} \gamma _2 z^{6}-\mathbf {i}w_n(\mathbf {i}w_n\tau _2+1)(\alpha _{2}+\delta _2\beta _{2})z^{4}-\gamma _2\kappa _{2}w_{n}^{2}z^{2}+\mathbf {i}w_n^3(\mathbf {i}w_n\tau _2+1)=0. \) 
In the sequel we denote, \(L_n:=\mathbf {i}w_n\tau _2+1\) , \(a:=\alpha _{2} \kappa _{2} \gamma _2,\)  \(b:=\alpha _{2}+\delta _2\beta _{2}\) , and \(c:=\gamma _2\kappa _{2}\) .
Lemma 3.8 
The equation
\(a z^{6}-\mathbf {i}w_nL_nbz^{4}-cw_{n}^{2}z^{2}+\mathbf {i}w_n^3L_n=0., \) 
admits six simple solutions, \(\pm z_{1},\pm z_{2}\)  and \(\pm z_{3}.\)

The solutions of (REF ) are the square roots of those of the following
equation
\(a X^{3}-\mathbf {i}w_nL_nbX^{2}-cw_{n}^{2}X+\mathbf {i}w_n^3L_n=0. \) 
By taking \(s=X-\mathbf {i}\frac{b}{3a}w_nL_n\) , that is \(X=s+\mathbf {i}\frac{b}{3a}w_nL_n\) 
the equation (REF ) becomes
\(s^{3}+ps+q=0 \) 
with
\(p &:=&\frac{w_n^2}{a}\left( \frac{(L_nb)^2}{3a}-c\right)=\frac{w_n^2L_n^2b^2}{3a^2}\left( 1+o(\frac{1}{w_n^{2-\varepsilon }})\right) \text{ and} \\q &:=&\mathbf {i}\frac{w_n^3L_n}{a}\left( \frac{2L_n^2b^3}{3^3a^2}+1-\frac{bc}{3a}\right)=\mathbf {i}2\frac{w_n^3L_nb^3}{3^3a^3}\left( 1+o(\frac{1}{w_n^{2-\varepsilon }}) \right)\) 
The discriminant of such equation is
\(\Delta &:=&q^{2}+\frac{4}{27}p^{3}=-M^2w_n^{6}L_n^4\left(1+o(\frac{1}{w_n^{2-\varepsilon }}) \right)\) 
where \(\varepsilon \)  is a small enough positive number and \(M^2:=\frac{4b^3}{3^3a^4}=\frac{2m}{a}\) .
Notice that for \(n\)  large enough, \(\Delta \ne 0\)  then, the equation (REF ) admits three simple solutions \(s_{1}=u_{0}+v_{0},\)  \(s_{2}=\mathbf {j}u_{0}+\mathbf {j}^{2}v_{0}\)  and \(s_{3}=\mathbf {j}^{2}u_{0}+\mathbf {j}v_{0}\)  with \(u_{0}^3=\frac{1}{2}(-q+\delta )\) , \(v_{0}^3=\frac{1}{2}(-q-\delta )\)  where \(\delta \)  is a square root of \(\Delta \) , \(\mathbf {j}=e^{\mathbf {i} \frac{2 \pi }{3}}\)  and such that \(u_0v_0=-\frac{p}{3}\) .
We take
\(\delta =\mathbf {i} Mw_n^{3}L_n^2\left(1+o(\frac{1}{w_n^{2-\varepsilon }}) \right)=\mathbf {i} Mw_n^{3}L_n^3\left(\frac{1}{L_n}+o(\frac{1}{w_n^{2-\varepsilon }}) \right).\) 
Then
\(u_0^3=-\mathbf {i}\frac{m}{2}w_n^3L_n^3\left(1-\frac{M}{mL_n}+o(\frac{1}{w_n^{2-\varepsilon }}) \right).\) 
and
\(v_0^3=-\mathbf {i}\frac{m}{2}w_n^3L_n^3\left(1+\frac{M}{mL_n}+o(\frac{1}{w_n^{2-\varepsilon }}) \right),\) 
where \(m:=\frac{2b^3}{3^3a^3}\) . We choose
\(u_0=\mathbf {i}\frac{b}{3a}w_nL_n\left(1-\frac{M}{3mL_n} +o(\frac{1}{w_n^{2-\varepsilon }}) \right),\) 
then
\(v_0=\mathbf {i}\frac{b}{3a}w_nL_n\left(1+\frac{M}{3mL_n} +o(\frac{1}{w_n^{2-\varepsilon }}) \right).\) 
Hence, the solutions of (REF ) are
\(\left\lbrace \begin{tabular}{l}s_{1} =\mathbf {i}\frac{2b}{3a}w_nL_n\left(1+o(\frac{1}{w_n^{2-\varepsilon }}) \right), \\s_{2} =-\mathbf {i}\frac{b}{3a}w_nL_n\left(1+\mathbf {i}\frac{\sqrt{3}M}{3mL_n}+o(\frac{1}{w_n^{2-\varepsilon }}) \right), \\s_{3} =-\mathbf {i}\frac{b}{3a}w_nL_n\left(1-\mathbf {i}\frac{\sqrt{3}M}{3mL_n}+o(\frac{1}{w_n^{2-\varepsilon }}) \right).\end{tabular}\right.\) 
The solutions of (REF ) are (\(x_1\)  associated to \(s_2\) )
\(\left\lbrace \begin{tabular}{l}x_{1} =\frac{bM}{3\sqrt{3}ma}w_n\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right) =b^{-1/2}w_n\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right), \\x_{2} =-\frac{bM}{3\sqrt{3}ma}w_n\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right) =-b^{-1/2}w_n\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right), \\x_{3} =\mathbf {i}\frac{b}{a}w_nL_n\left(1+o(\frac{1}{w_n^{2-\varepsilon }}) \right)=-\frac{b}{a}\tau _2 w_n^2\left(1-\mathbf {i}\frac{1}{\tau _2 w_n} +o(\frac{1}{w_n^{2-\varepsilon }})\right),\end{tabular}\right.\) 
they are simple and different from \(0.\)  Consequently, the equation (REF ) admits six simple solutions, \(\pm z_{1},\pm z_{2}\)  and \(\pm z_{3}\) , with
\(\left\lbrace \begin{tabular}{l}z_{1} =b^{-1/4}\sqrt{w_n}\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right) \\z_{2} =\mathbf {i}b^{-1/4}\sqrt{w_n}\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right), \\z_{3} =\mathbf {i}\sqrt{\frac{b\tau _2}{a}}w_n\left(1-\mathbf {i}\frac{1}{2\tau _2 w_n} +o(\frac{1}{w_n^{2-\varepsilon }}) \right).\end{tabular}\right.\) 
      \(\sqcup \unknown. {\sqcap }\)
Return back to the ODE (REF ). The general solution of (REF ) is given by
\(u_{2,n}=\sum _{k=1}^{3}(d_{k}e^{z_kx}+b_{k}e^{-z_kx})\) 
where \(d_{k},b_{k}\)  \(k=1,2,3\)  are some complex numbers dependent on \(n\) .
Finally, combining (REF ) and (REF ), it yields
\(\beta \theta _{2,n}=\mathbf {i}w_n\frac{\gamma _2\kappa _2}{\mathbf {i}w_n\tau _2+1}u_{2,n}-\delta _2\beta _2\partial _x^2u_{2,n}+\frac{\alpha _2\gamma _2\kappa _2}{\mathbf {i}w_n(\mathbf {i}w_n\tau _2+1)}\partial _x^4u_{2,n}\) 
which will allow us to deduce the explicit expression of \(\theta _{2,n}\) . Taking into account (REF ) and (REF ),
\(\beta _{2} \theta _{2,n}=\sum _{k=1}^{3}(-\frac{w_n^2}{z_{k}^{2}}+\alpha _{2}z_{k}^{2})(d_{k}e^{z_kx}+b_{k}e^{-z_{k}x}).\) 
We have also
\(\beta _{2} \partial _{x}\theta _{2,n}=\sum _{k=1}^{3}(-\frac{w_n^2}{z_{k}}+\alpha _{2} z_{k}^{3})(d_{k}e^{z_kx}-b_{k}e^{-z_{k}x})\) 
and
\(\beta _{2} \partial _{x}^{2}\theta _{2,n}=\sum _{k=1}^{3}(-w_n^2+\alpha _{2}z_{k}^{4})(d_{k}e^{z_kx}+b_{k}e^{-z_{k}x}).\) 
Now using the transmission and boundary conditions we have:
Lemma 3.9 
There exists a sequence \(\gamma _n:=\frac{w_{n}}{\sqrt{\alpha _{1}}}\)  of real positive numbers diverging to infinity such that
\(\frac{1}{w_n^{2\alpha }}\left( \left| -\frac{1}{2\gamma _{n}}+\gamma _{n}c_{1}\right| ^{2}+\left| \gamma _{n}c_{2}\right| ^{2}\right) \rightarrow +\infty . \) 

Step 1. We will built two sequences \(S_n\)  and \(T_n\)  such that
\(\gamma _{n}c_{1}(S_{n}-T_{n}\gamma _{n}\tan (\gamma _{n}\pi ))=\frac{1}{2\gamma _{n}}S_{n}-\frac{\pi }{2}\gamma _{n}T_{n} \) 
The transmission conditions at 0 are expressed as follow :
\(\sum _{k=1}^{3}(d_{k}+b_{k}) &=&\frac{1}{w_n^\alpha } c_{2},  \\\sum _{k=1}^{3}z_{k}(d_{k}-b_{k}) &=&0,  \\w_n^2\sum _{k=1}^{3}\frac{1}{z_{k}}(d_{k}-b_{k}) &=&\frac{\alpha _1}{w_n^\alpha }\left( -\frac{1}{2\gamma _{n}}+\gamma _{n}c_{1}\right) ,  \\\sum _{k=1}^{3}(-\frac{w_n^2}{z_{k}^{2}}+\alpha _{2} z_{k}^{2})(d_{k}+b_{k}) &=&0,\) 
and the boundary conditions at \(\ell _{1}\)  and \(\ell _{2}\)  are
\(\sum _{k=1}^{3}(d_{k}e^{z_{k}\pi }+b_{k}e^{-z_{k}\pi }) &=&0,  \\\sum _{k=1}^{3}z_{k}^{2}(d_{k}e^{z_{k}\pi }+b_{k}e^{-z_{k}\pi }) &=&0,  \\\sum _{k=1}^{3}\frac{1}{z_{k}^{2}}(d_{k}e^{z_{k}\pi }+b_{k}e^{-z_{k}\pi }) &=&0, \\c_{1}\sin (\gamma _{n}\pi )+(-\frac{\pi }{2\gamma _{n}}+c_{2})\cos (\gamma _{n}\pi ) &=&0.\) 
Taking \(s_{k}:=d_{k}+b_{k}\)  and \(t_{k}:=d_{k}-b_{k}.\)  Then (REF ), () and (), () can be rewritten as follow
\(s_{1}+s_{2}+s_{3} &=&\frac{1}{w_n^\alpha } c_{2},  \\(-\frac{w_n^2}{z_{1}^{2}}+\alpha _{2} z_{1}^{2})s_{1}+(-\frac{w_n^2}{z_{2}^{2}}+\alpha _{2}z_{2}^{2})s_{2}+(-\frac{w_n^2}{z_{3}^{2}}+\alpha _{2} z_{3}^{2})s_{3} &=&0\) 
and
\(z_{1}t_{1}+z_{2}t_{2}+z_{3}t_{3} &=&0,  \\\frac{1}{z_{1}}t_{1}+\frac{1}{z_{2}}t_{2}+\frac{1}{z_{3}}t_{3}&=&\frac{1}{w_{n}^{2+\alpha }} \left( -\frac{1}{2\gamma _{n}}+\gamma _{n}c_{1}\right) =:-T. \) 
It yields
\(X_{1}t_{1}+X_{2}t_{2} &=&z_{3}^{2}T, \\Z_{1}s_{1}+Z_{2}s_{2} &=&Z_{3}c_{2}\) 
with
\(\left\lbrace \begin{tabular}{l}X_{j} :=z_{j}-\frac{z_{3}^{2}}{z_{j}}=\frac{1}{z_{j}}(z_{j}^{2}-z_{3}^{2}),\;\;j=1,2, \\Z_{j} :=\alpha _{2} (z_{j}^{2}-z_{3}^{2})-w_n^2(\frac{1}{z_{j}^{2}}-\frac{1}{z_{3}^{2}})=(\alpha _{2} +\frac{w_n^2}{z_{j}^{2}z_{3}^{2}})(z_{j}^{2}-z_{3}^{2}),\;\;j=1,2, \\Z_{3}:=\frac{w_n^2}{z_{3}^{2}}-\alpha _{2} z_{3}^{2}.\end{tabular}\right.\) 
We deduce that
\(2d_{2} &=&a_{1}s_{1}+a_{2}t_{1}+a_{3}c_{2}+a_{4}T,  \\2b_{2} &=&a_{1}s_{1}-a_{2}t_{1}+a_{3}c_{2}-a_{4}T \) 
with
\(a_{1}:=-\frac{Z_{1}}{Z_{2}},\;a_{2}:=-\frac{X_{1}}{X_{2}},\;a_{3}:=\frac{Z_{3}}{Z_{2}}\;\text{and }a_{4}:=\frac{z_{3}^{2}}{X_{2}}. \) 
Return back to (REF ) and (REF ), we obtain
\(2d_{3} &=&A_{1}s_{1}+A_{2}t_{1}+A_{3}c_{2}+A_{4}T,  \\2b_{3} &=&A_{1}s_{1}-A_{2}t_{1}+A_{3}c_{2}-A_{4}T \) 
with
\(\left\lbrace \begin{tabular}{l}A_{1} =-1-a_1=\frac{Z_{1}}{Z_{2}}-1=(\alpha +\frac{w_n^2}{z_{1}^{2}z_{2}^{2}})\frac{z_{1}^{2}-z_{2}^{2}}{Z_{2}}, \\A_{2} =\frac{z_{2}}{z_{3}}\frac{X_{1}}{X_{2}}-\frac{z_{1}}{z_{3}}=\frac{z_{3}(z_{1}^{2}-z_{2}^{2})}{z_{1}(z_{2}^{2}-z_{3}^{2})}, \\A_{3} =1-a_3=1-\frac{Z_{3}}{Z_{2}}=-\frac{w_n^2-\alpha z_{2}^{4}}{z_{2}^{2}Z_{2}}, \\A_{4} =-\frac{z_2}{z_3}a_4=-\frac{z_{2}z_{3}}{X_{2}}=-\frac{z_{2}^{2}z_{3}}{z_{2}^{2}-z_{3}^{2}}.\end{tabular}\right.\) 
Replacing \(b_{2},d_{2},b_{3}\)  and \(d_{3}\)  given by (REF )-() in (REF )-(), we obtain
\(f_{j}d_{1}+g_{j}b_{1}+h_{j}c_{2}+l_{j}T=0,\;\;j=0,2,-2 \) 
with
\(\left\lbrace \begin{tabular}{l}f_{j} :=2z_{1}^{j}e^{z_{1}\pi }+z_{2}^{j}(a_{1}+a_{2})e^{z_{2}\pi }+z_{3}^{j}(A_{1}+A_{2})e^{z_{3}\pi } \\\;\;\;\;\;\;\;\;\;+z_{2}^{j}(a_{1}-a_{2})e^{-z_{2}\pi }+ z_{3}^{j}(A_{1}-A_{2})e^{-z_{3}\pi }, \\\;\;\;\;\;=e^{z_{1}\pi }(2z_{1}^{j}+e^{-\varepsilon \sqrt{w_n}}o(1)),\\g_{j} :=z_{2}^{j}(a_{1}-a_{2})e^{z_{2}\pi }+ z_{3}^{j}(A_{1}-A_{2})e^{z_{3}\pi } \\\;\;\;\;\;\;\;\;\;+2z_{1}^{j}e^{-z_{1}\pi }+z_{2}^{j}(a_{1}+a_{2})e^{-z_{2}\pi }+z_{3}^{j}(A_{1}+A_{2})e^{-z_{3}\pi }, \\h_{j} :=a_{3}z_{2}^{j}e^{z_{2}\pi }+ A_{3}z_{3}^{j}e^{z_{3}\pi } +a_{3}z_{2}^{j}e^{-z_{2}\pi }+A_{3}z_{3}^{j}e^{-z_{3}\pi }, \\l_{j} :=a_{4}z_{2}^{j}e^{z_{2}\pi }+ A_{4}z_{3}^{j}e^{z_{3}\pi } -a_{4}z_{2}^{j}e^{-z_{2}\pi }-A_{4}z_{3}^{j}e^{-z_{3}\pi }.\end{tabular}\right.\) 
Combining () and () with (REF ), it follows
\(w_{n}^{2}(f_{j}d_{1}+g_{j}b_{1})=\left( -\frac{\pi }{2\gamma _{n}}+c_{1}\tan (\gamma _{n}\pi )\right) h_{j}+\frac{\alpha _1}{w_n^\alpha }\left( -\frac{1}{2\gamma _{n}}+\gamma _{n}c_{1}\right)l_{j},\;\;j=0,2. \) 
After some combinations we conclude, using (REF ), the following
equation,
\(\alpha _1\left( -\frac{1}{2\gamma _{n}}+\gamma _{n}c_{1}\right) S_{n}=w_{n}^{2}\left( -\frac{\pi }{2\gamma _{n}}+c_{1}\tan (\gamma _{n}\pi )\right) T_{n} \) 
where
\(&&S_{n}:=(g_{2}f_{0}-g_{0}f_{2})l_{-2}-(g_{2}l_{0}-g_{0}l_{2})f_{-2}+(f_{2}l_{0}-f_{0}l_{2})g_{-2},\\&&T_{n}:=-(g_{2}f_{0}-g_{0}f_{2})h_{-2}+(g_{2}h_{0}-g_{0}h_{2})f_{-2}-(f_{2}h_{0}-f_{0}h_{2})g_{-2}.\) 
From (REF ) one has
\(\gamma _{n}c_{1}(S_{n}-T_{n}\gamma _{n}\tan (\gamma _{n}\pi ))=\frac{1}{2\gamma _{n}}S_{n}-\frac{\pi }{2}\gamma _{n}T_{n}\) 
For \(k:=\frac{b\pi }{a\kappa _2}\) , \(\theta _2=\frac{1}{\mathbf {i}}z_2\pi \) , \(\theta _3=\frac{1}{\mathbf {i}}(z_3\pi -k)\)  and \(P:=z_{1}^{-2}(z_{2}^{2}-z_{3}^{2})+z_{2}^{-2}(z_{3}^{2}-z_{1}^{2})+z_{3}^{-2}(z_{1}^{2}-z_{2}^{2}),\)  and by using (REF ), we have the following expressions of \(S_n\)  and \(T_n\) ,
\(S_{n} &=&P\left( D_{1}e^ke^{\mathbf {i}(\theta _2+\theta _3)}+D_{2}e^{-k}e^{-\mathbf {i}(\theta _2+\theta _3)}+D_{3}e^{-k}e^{\mathbf {i}(\theta _2-\theta _3)}+D_{4}e^ke^{-\mathbf {i}(\theta _2-\theta _3)}+e^{-\varepsilon \sqrt{w_n}}o(1)\right)\\&=&P\left[\left( (D_1+D_4)e^k+(D_2+D_3)e^{-k}\right) \cos (\theta _2)\cos (\theta _3)\right.\\&-&\left( (D_1-D_4)e^k+(D_2-D_3)e^{-k}\right)\sin (\theta _2)\sin (\theta _3)\\&+&\mathbf {i}\left( (D_1-D_4)e^k-(D_2-D_3)e^{-k}\right) \sin (\theta _2)\cos (\theta _3)\\&+&\mathbf {i}\left( (D_1+D_4)e^k-(D_2+D_3)e^{-k}\right)\cos (\theta _2)\sin (\theta _3)\\&+&\left.e^{-\varepsilon \sqrt{w_n}}o(1)\right].\) 
and
\(T_{n} &=&P\left( D^\prime _{1}e^ke^{\mathbf {i}(\theta _2+\theta _3)}+D^\prime _{2}e^{-k}e^{-\mathbf {i}(\theta _2+\theta _3)}+D^\prime _{3}e^{-k}e^{\mathbf {i}(\theta _2-\theta _3)}+D^\prime _{4}e^ke^{-\mathbf {i}(\theta _2-\theta _3)}+e^{-\varepsilon \sqrt{w_n}}o(1)\right)\\&=&P\left[\left( (D^\prime _1+D^\prime _4)e^k+(D^\prime _2+D^\prime _3)e^{-k}\right) \cos (\theta _2)\cos (\theta _3)\right.\\&-&\left( (D^\prime _1-D^\prime _4)e^k+(D^\prime _2-D^\prime _3)e^{-k}\right)\sin (\theta _2)\sin (\theta _3)\\&+&\mathbf {i}\left( (D^\prime _1-D^\prime _4)e^k-(D^\prime _2-D^\prime _3)e^{-k}\right) \sin (\theta _2)\cos (\theta _3)\\&+&\mathbf {i}\left( (D^\prime _1+D^\prime _4)e^k-(D^\prime _2+D^\prime _3)e^{-k}\right)\cos (\theta _2)\sin (\theta _3)\\&+&\left.e^{-\varepsilon \sqrt{w_n}}o(1)\right].\) 
where \(\varepsilon \)  is a small enough positive number and with
\(&&D_{1} :=bA_4-a_4B,\;\; D_{2} :=a_4A-aA_4,\\&&D_{3} :=-(bA_4+a_4A),\;\;D_{4} :=aA_4+Ba_4 \\\) 
and
\(&&D^\prime _{1} :=bA_3-a_3B,\;\;D^\prime _{2} :=aA_3-a_3A, \\&&D^\prime _{3} :=bA_3-a_3A,\;\;D^\prime _{4} :=aA_3-Ba_3. \\\) 
Step 2. Estimate of \(S_n\)  and \(T_n\)  and conclusion.
We focus now on the asymptotic behavior of \(S_{n}\)  and \(T_{n}\) 
under \(n\) . To do this, we need the estimates of \(a_i, \;\;i=1,\cdots ,4\)  and \(A_i, \;\;i=1,\cdots ,4\) .
It remains first to estimate \(Z_1, Z_2,Z_3\)  and \(X_1, X_2\) . Using (REF ) in (REF ), we get
\(Z_1 =\frac{b}{a}\alpha _2\tau _2 w_n^2\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right), &Z_2 =\frac{b}{a}\alpha _2\tau _2 w_n^2\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right), &Z_3 =\frac{b}{a}\alpha _2\tau _2 w_n^2\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right),\) 
\(X_1 &=\frac{b^{5/4}}{a}\tau _2 w_n^{3/2}\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right),&X_2 &=-\mathbf {i} \frac{b^{5/4}}{a}\tau _2 w_n^{3/2}\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right).\) 
So by (REF ) and (REF ), we obtain
\(a_1 &=-\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right), &a_2 &=-\mathbf {i} \left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right),\\a_3&=\left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right),&a_4&=-\mathbf {i}\frac{1}{b^{1/4}}\sqrt{w_n} \left(1++o(\frac{1}{w_n^{1-\varepsilon }}) \right)\) 

and
\(A_1 &=\frac{1}{\sqrt{w_n}} o(\frac{1}{w_n^{1/2-\varepsilon }}),&A_2&=2\mathbf {i}\frac{a^{1/2}}{b^{3/4}\tau _2^{1/2}}\frac{1}{\sqrt{w_n}} \left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right),\\A_3 &=o(\frac{1}{w_n^{1-\varepsilon }}),&A_4&=\mathbf {i}\frac{a^{1/2}}{b\tau _2^{1/2}} \left(1+o(\frac{1}{w_n^{1-\varepsilon }}) \right).\) 
So, we have the following estimates of \(D_1\pm D_4\) , \(D_2\pm D_3\) , \(D_1^\prime \pm D_4^\prime \)  and \(D_2^\prime \pm D_3^\prime \) ,
\(D_1+D_4 &=-2\mathbf {i}\frac{a^{1/2}}{b\tau _2^{1/2}}\left(1 +o(\frac{1}{w_n^{1/2-\varepsilon }})\right), &D_1-D_4&=-2\frac{a^{1/2}}{b\tau _2^{1/2}}\left(1+o(\frac{1}{w_n^{1/2-\varepsilon }})\right),&\\D_2+D_3 &=-(D_1+D_4),&D_2-D_3&=-2\frac{a^{1/2}}{b\tau _2^{1/2}}\left(1+o(\frac{1}{w_n^{1/2-\varepsilon }})\right).\) 
and
\(D^\prime _1+D^\prime _4 &=4\mathbf {i}\frac{a^{1/2}}{b^{3/4}\tau _2^{1/2}}\frac{1}{\sqrt{w_n}}\left(1 +o(\frac{1}{w_n^{1/2-\varepsilon }})\right),&D^\prime _1-D^\prime _4 &=-\frac{1}{\sqrt{w_n}}o(\frac{1}{w_n^{1/2-\varepsilon }}),&\\D^\prime _2+D^\prime _3 &=-4\mathbf {i}\frac{a^{1/2}}{b^{3/4}\tau _2^{1/2}}\frac{1}{\sqrt{w_n}}\left(1 +o(\frac{1}{w_n^{1/2-\varepsilon }})\right),&D^\prime _2-D^\prime _3&=-\left(D^\prime _1-D^\prime _4 \right).\) 

Hence, taking in mind that \(P\ne 0\)  (by Vieta's formula associated to equation (REF )),
\(\frac{1}{P} S_{n} &=&-4\mathbf {i}\frac{a^{1/2}}{b\tau _2^{1/2}}\left( \sinh (k)\cos \theta _3+\mathbf {i}\cosh (k)\sin \theta _3\right) \left(1+o(\frac{1}{w_n^{1/2-\varepsilon }})\right)\cos \theta _2\\&-&4\mathbf {i}\frac{a^{1/2}}{b\tau _2^{1/2}}\left(\sinh (k)\cos \theta _3+\mathbf {i} \cosh (k)\sin \theta _3\right) \left(1+o(\frac{1}{w_n^{1/2-\varepsilon }})\right)\sin \theta _2+o(\frac{1}{w_n^{1/2-\varepsilon }})\\&=&4\mathbf {i}\frac{a^{1/2}}{b\tau _2^{1/2}}\left( \sinh (k)\cos \theta _3+\mathbf {i}\cosh (k)\sin \theta _3\right) \left(1+o(\frac{1}{w_n^{1/2-\varepsilon }})\right)\left( \cos \theta _2+\sin \theta _2\right)+o(\frac{1}{w_n^{1/2-\varepsilon }})\) 
and
\(\frac{1}{P} T_{n} =&\frac{1}{\sqrt{ w_n}}\left[ 8\mathbf {i}\frac{a^{1/2}}{b^{3/4}\tau _2^{1/2}}\left( \sinh (k)\cos \theta _3+\mathbf {i}\cosh (k)\sin \theta _3\right) \left(1 +o(\frac{1}{w_n^{1/2-\varepsilon }})\right)\cos \theta _2+ o(\frac{1}{w_n^{1/2-\varepsilon }})\right].\) 
We take \(s_n:=\frac{1}{\sqrt{b}}w_n=\sqrt{\frac{\alpha _1}{b}}\gamma _n\) .
By using the Asymptotic Dirichlet's theorem [2]}, there exists \((p_{n},q_{n})\in \mathbb {N}^{2}\)  such that \(p_n,q_n\rightarrow \infty \)  and
\(\left| \left(\frac{\alpha _1}{b} \right)^{1/4} -\frac{p_n}{q_n}\right| <\frac{1}{q_{n}^2}\) 
which implies
\(\left| \left(\frac{\alpha _1}{b} \right)^{1/4}q_n-p_n\right| <\frac{1}{q_n}. \) 
One has \(\sqrt{\frac{b}{a}}w_n=\frac{b}{\sqrt{a}}s_n\)  and \(\gamma _n=\sqrt{\frac{b}{\alpha _1}}s_n\) .
We choose
\(\sqrt{\gamma _n}=q_n+\frac{\alpha }{q_n^2},\) 
then
\(\sqrt{s_n}=\left(\frac{\alpha _1}{b} \right)^{1/4}q_n+\left(\frac{\alpha _1}{b} \right)^{1/4}\frac{\alpha }{q_n^2}.\) 
Recall that
\(\frac{\theta _2}{\pi }=\sqrt{s_n} +o(\frac{1}{s_n^{1/2-\varepsilon }}), \) 
then
\(\sin (\theta _2) = o(\frac{1}{q_n^{1-\varepsilon }})\;\;\;\text{and}\;\;\;\cos (\theta _2) = 1+o(\frac{1}{q_n^{1-\varepsilon }}).\) 
Moreover,
\(\tan (\gamma _n\pi )=\frac{2\alpha }{q_n}\left( 1+o(\frac{1}{q_n^{1-\varepsilon }})\right). \) 
Hence
\(\frac{1}{P}S_n=\mathbf {i}\frac{a^{1/2}}{b\tau _2^{1/2}}\left( \sinh (k)\cos \theta _3+\mathbf {i}\cosh (k)\sin \theta _3\right) \left(1+o(\frac{1}{q_n^{1-\varepsilon }})\right) \cos \theta _2+o(\frac{1}{q_n^{1-\varepsilon }})\) 
and
\(&&\frac{1}{P}T_{n}\gamma _{n}\tan (\gamma _{n}\pi )\\&&=2\alpha \alpha _1^{-1/4}8\mathbf {i}\frac{a^{1/2}}{b^{3/4}\tau _2^{1/2}}\left( \sinh (k)\cos \theta _3+\mathbf {i}\cosh (k)\sin \theta _3\right) \left(1 +o(\frac{1}{q_n^{1-\varepsilon }})\right)\cos \theta _2+ o(\frac{1}{q_n^{1-\varepsilon }}).\) 
We choose \(\alpha \)  such that \(S_{n}-\alpha _{1}T_{n}\gamma _{n}\tan (\gamma _{n}\pi )=o(\frac{1}{q_n^{1-\varepsilon }})\)  i.e. \(\alpha =\frac{\alpha _1^{1/4}}{4b^{1/4}}\) ,
to conclude that
\(|\gamma _nc_1|\ge C \gamma _n^{1-\varepsilon }\) 
for some positive constant \(C\) .
Thus we have
\(\frac{1}{w_n^{2\alpha }}\left( \left| -\frac{1}{2\gamma _{n}}+\gamma _{n}c_{1}\right| ^{2}+\left| \gamma _{n}c_{2}\right| ^{2}\right) \rightarrow +\infty .\) 
The proof of Lemma REF  is thus complete.      \(\sqcup \unknown. {\sqcap }\)
Finally, taking the real part of the inner product of (REF ) with \((\pi -x)\partial _{x}u_{1,n}\)  to obtain
\(&&-\frac{\pi w_n^{\alpha }}{2}\left( \left| -\frac{1}{2\gamma _{n}} +\gamma _{n}c_{1}\right| ^{2}+\left| \gamma _{n}c_{2}\right| ^{2}\right)\\&&=-\frac{w_n^{\alpha }}{2}\left( \gamma _{n}^{2}\left\Vert  u_{1,n}\right\Vert  ^{2}+\left\Vert  \partial _{x}u_{1,n}\right\Vert ^{2}\right) +Re(\int _{0}^{\pi }\alpha _{1}\sin (\gamma _{n}x)(\pi -x)\partial _{x}\overline{u_{1,n}}dx).\) 
Then, By (REF ), \(\gamma _{n}^{2}\left\Vert  u_{1,n}\right\Vert  ^{2}+\left\Vert  \partial _{x}u_{1,n}\right\Vert ^{2}\)  diverges to infinity. In conclusion \(y_{n}\)  is not bounded.      \(\sqcup \unknown. {\sqcap }\)
